- You can eliminate Nr9 in r6c3 because it is already in r7c3 and r9c3 ( must be a 9 in the 3 x 3 square ).
- This leads to a pair of 6,9 in r4c1 and r4c2. Therefore, r4c7 must be Nr 3 and r4c5 must be Nr 4
- This leads to a pair of 1,4 in r5c2 and r5c3.
- Eventually, leads to a pair of 2,4 in r8c1 and r8c2 which means r8c3 must be Nr1.